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September 30th, 2014, 06:53 AM  #1 
Member Joined: Mar 2013 Posts: 71 Thanks: 4  How to prove that $2$ is an irreducible element in $\mathbb {Z}[\sqrt{5}]$?
How to prove that $2$ is an irreducible element in $\mathbb {Z}[\sqrt{5}]$? One way is to check if there exists $a,b,c,d \in \mathbb Z$ such that $2=(a+b\sqrt{5})(c+d\sqrt{5})$, i.e., $2=ac+ 5bd+ \sqrt{5}(ad+bc)$. It leads to the system $ac+5bd=2$ and $ad+bc=0$, which is difficult to solve. Is there another way? Thanks!

October 4th, 2014, 02:08 AM  #2 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88 
For $a + b\sqrt{5} \in \Bbb Z[\sqrt{5}]$ define its norm $N$ by: $N(a + b\sqrt{5}) = a^2  5b^2$. Note that: $N((a + b\sqrt{5})(c + d\sqrt{5})) = N((ac + 5bd) + (ad + bc)\sqrt{5}) = (ac + 5bd)^2  5(ad + bc)^2$ $=a^2c^2 + 10abcd + 25b^2d^2  5a^2d^2  10abcd  5b^2c^2$ $= a^2c^2  5b^2c^2  5a^2d^2 + 25b^2d^2 = (a^2  5b^2)(c^2  5d^2) = N(a + b\sqrt{5})N(c + d\sqrt{5})$. So $N$ is multiplicative. Clearly, if $a + b\sqrt{5}$ is a unit, then, we must have $N(a + b\sqrt{5}) = \pm 1$. Suppose, for the sake of argument, $N(a + b\sqrt{5}) = 1$, so: $1 = 5b^2  a^2$. If $a \geq b$, then: $1 = 5b^2  a^2 \geq 4b^2$. This is a contradiction, unless $b = 0$, in which case we have: $1 = a^2$, which is impossible. Similarly, if $a < b$, then: $1 = 5b^2  a^2 > 4a^2$ which forces $a = 0$, leading to a similar contradiction. Thus if $a + b\sqrt{5}$ is a unit, it has norm 1. The converse is clear: if $N(a + b\sqrt{5}) = 1$, then $a^2  5b^2 = 1$, in which case: $a + b\sqrt{5}$ has inverse $a  b\sqrt{5}$. For example, $9 + 4\sqrt{5}$ is a unit. Now if $2 = (a + b\sqrt{5})(c + d\sqrt{5})$, then: $4 = (a^2  5b^2)(c^2  5d^2)$. As we saw above, NO element of $\Bbb Z[\sqrt{5}]$ has norm 1, and if one of the factors of 2 has norm 1, it is a unit, so it doesn't count (a ring element is reducible only if it factors into two nonunits). This means if 2 is reducible, its nonunit factors must each have norm $\pm 2$. The same argument we used to show no element of $\Bbb Z[\sqrt{5}]$ has norm 1 also works to show that no element has norm 2. So we must have $a^2 = 5b^2 + 2$. Here, we can use a clever trick: $a^2 = 2$ (mod 5). Checking, we see that: $0^2 = 0$ (mod 5) $1^2 = 1$ (mod 5) $2^2 = 4$ (mod 5) $3^2 = 4$ (mod 5) $4^2 = 1$ (mod 5), so there is no solution. (If you prefer, write $a = 5k + n$ for $n = 0,1,2,3,4$, and put all the multiples of 5 on one side of the equation). My apologies for the length of this post. 

Tags 
$2$, $mathbb, element, irreducible, prove, zsqrt5$ 
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